Problems

Solution: \begin{questionparts} \item \(\,\) \begin{align*} && I_n &= \int_0^{\beta} (\sec x + \tan x)^n \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} \left ( (\sec x + \tan x)^{n+1}+(\sec x + \tan x)^{n-1}\right) \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( (\sec x + \tan x)^{2}+1\right) \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( \sec^2 x + \tan^2 x + 2\sec x \tan x + 1\right) \, \d x \\ && \tfrac12(I_{n+1}+I_{n-1}) &= \tfrac12\int_0^{\beta} (\sec x + \tan x)^{n-1}\left ( 2\sec x \tan x +2\sec^2 x \right) \, \d x \\ &&& = \left [\frac1n(\sec x + \tan x)^{n} \right]_0^{\beta} \\ &&&= \frac1n[(\sec \beta + \tan \beta)^n - 1] \end{align*} Notice that by AM-GM \(\tfrac12( ( (\sec x + \tan x)^{n+1}+(\sec x + \tan x)^{n-1}) \geq (\sec x + \tan x)^{n}\) with equality not holding most of the time. Integrating we obtain our result. \item \(\,\) \begin{align*} && J_n &= \int_0^{\beta} (\sec x \cos \beta + \tan x )^n \d x \\ && \tfrac12( J_{n+1} + J_{n-1}) &= \tfrac12 \int_0^{\beta} \left ( (\sec x \cos \beta + \tan x )^{n+1} +(\sec x \cos \beta + \tan x )^{n-1}\right ) \d x \\ && &= \tfrac12 \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( (\sec x \cos \beta + \tan x )^{2} + \right ) \d x \\ && &= \tfrac12 \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec^2 x \cos^2 \beta + \tan^2 x+ 2\sec x \tan x \cos \beta +1 \right ) \d x \\ && &= \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec x \tan x \cos \beta +\tfrac12(\cos^2 \beta +1)\sec^2 x \right ) \d x \\ && &< \int_0^{\beta}(\sec x \cos \beta + \tan x )^{n-1} \left ( \sec x \tan x \cos \beta +\sec^2 x \right ) \d x \\ &&&= \left [\frac1n (\sec x \cos \beta + \tan x)^{n} \right]_0^{\beta} \\ &&&= \frac1n[ (1 + \tan \beta)^n - \cos^n \beta] \end{align*} But notice we can use the same AM-GM argument from before to show that \(J_n < \tfrac12( J_{n+1} + J_{n-1}) < \frac1n[ (1 + \tan \beta)^n - \cos^n \beta]\)